Linear Algebra Exercises-n-Answers.pdf

140 Linear Algebra, by Hefferon
and so (as in Example 3.5 and 3.6, we can just find the vectors perpendicular to all of the members
of the basis)
( ) ( ) (
M ⊥ u ∣∣ 1 ∣∣ 1
= { −1 · u + 1 · v = 0} = {k · k ∈ R} B
v
1
M ⊥ = 〈 〉
1)
and representing the vector with respect
(
to the concatenation
( ) (
gives this.
1 −1 1
= −2 · − 1 ·
−3)
1 1)
Keeping the M part yields the answer.
( (
1 2
proj M,M ⊥( ) =
−3)
−2)
The third part is also a simple calculation (there is a 1×1 matrix in the middle, and the inverse
of it is also 1×1)
A ( A trans A ) ( ) (
−1
A trans −1 (−1 ) ( )) −1 ( )
−1 ( ) −1 (2 ) −1 ( )
=
1 −1 1 = −1 1
1
1
1
( ) ( ) ( )
−1 (1/2 ) ( ) −1 (−1/2 ) 1/2 −1/2
=
−1 1 = 1/2 =
1
1
−1/2 1/2
which of course gives the same answer.
( ( ) ( (
1 1/2 −1/2 1 2
proj M ( ) =
=
−3)
−1/2 1/2 −3)
−2)
(b) Paramatrization gives this.
⎛
M = {c · ⎝ −1 ⎞
0 ⎠ ∣ c ∈ R}
1
With that, the formula for
⎛
the first way gives this.
⎝ 0 ⎞ ⎛
1⎠
⎝ −1
⎞
0 ⎠ ⎛
2 1
⎛
⎝ −1
⎞ ⎛
0 ⎠ ⎝ −1
⎞ · ⎝ −1
⎞ ⎛
0 ⎠ = 2 2 · ⎝ −1
⎞ ⎛
0 ⎠ = ⎝ −1
⎞
0 ⎠
1
1 1
0 ⎠
1 1
To proceed by the second method
⎛ ⎞
we find M ⊥ ,
⎛ ⎞ ⎛ ⎞
M ⊥ = { ⎝ u v ⎠ ∣ −u + w = 0} = {j · ⎝ 1 0⎠ + k · ⎝ 0 1⎠ ∣ j, k ∈ R}
w
1 0
find the representation of the given vector with respect to the concatenation of the bases B M and
B M ⊥
⎛
⎝ 0 ⎞ ⎛
1⎠ = 1 · ⎝ −1
⎞ ⎛
0 ⎠ + 1 · ⎝ 1 ⎞ ⎛
0⎠ + 1 · ⎝ 0 ⎞
1⎠
2 1 1 0
and retain only the M part.
⎛
proj M ( ⎝ 0 ⎞ ⎛
1⎠) = 1 · ⎝ −1
⎞ ⎛
0 ⎠ = ⎝ −1
⎞
0 ⎠
2
1 1
Finally, for the third method, the matrix calculation
⎛ ⎞
⎛ ⎞
⎛ ⎞
A ( A trans A ) −1
−1
A trans = ⎝ 0 ⎠ (( −1 0 1 ) −1
⎝ 0 ⎠ ) −1 ( )
−1
−1 0 1 = ⎝ 0 ⎠ ( 2 ) −1 ( )
−1 0 1
1
1
1
⎛
= ⎝ −1 ⎞
⎛
0 ⎠ ( 1/2 ) ( −1 0 1 ) = ⎝ −1 ⎞
⎛
⎞
0 ⎠ ( −1/2 0 1/2 ) 1/2 0 −1/2
= ⎝ 0 0 0 ⎠
1
1
−1/2 0 1/2
followed by matrix-vector multiplication
⎛
proj M ( ⎝ 0 ⎞ ⎛
⎞ ⎛
1/2 0 −1/2
1⎠)
⎝ 0 0 0 ⎠ ⎝ 0 ⎞ ⎛
1⎠ = ⎝ −1 ⎞
0 ⎠
2 −1/2 0 1/2 2 1
gives the answer.